how can I determine the domain of a function h(x): integrate from 0 to 2x-1 f(t)dt, when the f(x) function is a graph shown and equations for both semicircles in the graph are: square root of 4x-x(squared) and for the second one -square root(1-(x-5)squared). later find derivative of function h(x) when evaluated at 5/2
To determine the domain of the function h(x) = ∫[0 to 2x-1] f(t) dt, we need to examine the two provided semicircle functions, √(4x-x^2) and -√(1-(x-5)^2), that make up the integrand.
The domain of a function is the set of all possible input values that the function can accept. In this case, we need to consider any restrictions on x that might exist within the given semicircles.
For the first semicircle, √(4x-x^2), we need to find the range of x-values where the expression inside the square root is non-negative. Since the square root of a negative number is undefined, we must have 4x-x^2 ≥ 0. To solve this quadratic inequality, we can factor it as (x-2)(x-2) ≥ 0, which simplifies to (x-2)^2 ≥ 0. This quadratic function is always non-negative for any value of x, so there are no restrictions on the domain for the first semicircle: (-∞, +∞).
For the second semicircle, -√(1-(x-5)^2), we need to find the range of x-values where the expression inside the square root is defined. In this case, (x-5)^2 ≤ 1, since the square root of a negative number or a number larger than 1 is undefined. We can solve this inequality by taking the square root of both sides, resulting in |x-5| ≤ 1. This means that x must lie within 4 and 6, including both endpoints: [4, 6].
Now that we have determined the domain of each semicircle, we can find the intersection of the two domains to determine the overall domain for h(x). Since the first semicircle has no restrictions, its domain (-∞, +∞) intersects with the restricted domain of the second semicircle [4, 6]. Therefore, the domain of h(x) is [4, 6].
To find the derivative of h(x) when evaluated at 5/2, we can apply the Fundamental Theorem of Calculus. According to this theorem, if a function h(x) is defined as the integral (with upper and lower limits) of another function, then the derivative of h(x) can be found by taking the derivative of the integrand and evaluating it at the given limits.
In this case, we have h(x) = ∫[0 to 2x-1] f(t) dt. To find the derivative of h(x), let's denote the integrand as F(t) (since f(t) is already used for the semicircles). Now we can calculate h'(x) = d/dx ∫[0 to 2x-1] F(t) dt using the chain rule.
Applying the chain rule, we have h'(x) = F(2x-1) * d/dx(2x-1). Now, we need to evaluate this derivative at x = 5/2.
To do that, we first evaluate F(2x-1) at x = 5/2. Substitute x = 5/2 into F(t), which represents the function being integrated, based on the provided semicircles.
Finally, we differentiate (d/dx) (2x-1) to find the derivative of the upper limit of the integral. This derivative simplifies to 2.
By multiplying F(2x-1) evaluated at x = 5/2 with the derivative of the upper limit, we can find h'(x) when evaluated at x = 5/2.
To determine the domain of the function h(x) = ∫(0 to 2x-1) f(t) dt, given the graphs of the semicircles as f(x), we need to consider the range of x-values where the integrals are defined.
Let's analyze the equations of each semicircle separately:
1) The equation of the first semicircle is √(4x - x^2). To determine its domain, we need to find the values of x for which the expression inside the square root is non-negative.
4x - x^2 ≥ 0
Factoring out x, we get:
x(4 - x) ≥ 0
The critical points are x = 0 and x = 4. We can use these points to test for the sign of the expression. For x < 0 or x > 4, both factors will be negative or positive, so the expression is positive. For the interval 0 < x < 4, one factor is negative and the other is positive, so the expression is negative. Therefore, the domain of the first semicircle is [0, 4].
2) The equation of the second semicircle is -√(1 - (x - 5)^2). Here again, we need to find the values of x for which the expression inside the square root is non-negative.
1 - (x - 5)^2 ≥ 0
Expanding the square, we get:
1 - (x^2 - 10x + 25) ≥ 0
- x^2 + 10x - 24 ≥ 0
The critical points are x = 4 and x = 6. By testing each interval, we find that the expression is negative for 4 < x < 6, and positive otherwise. Therefore, the domain of the second semicircle is [4, 6].
To find the domain of the function h(x), we need both semicircles to be defined. Thus, the common domain for both semicircles is [4, 6]. Therefore, the domain of h(x) is [4, 6].
Now, let's find the derivative of h(x) when evaluated at 5/2. Using the Fundamental Theorem of Calculus, the derivative of h(x) is simply f(2x-1). In this case, we need to evaluate f(t) at t = 2(5/2) - 1 = 5 - 1 = 4.
Plugging x = 5/2 into the first semicircle equation, we find:
f(4) = √(4(4) - (4^2)) = √(16 - 16) = √0 = 0
Therefore, the derivative of h(x) when evaluated at 5/2 is 0.